I can't remove the _id index, why?
When I try running the dropIndexes command, it removes all indexes but not the _id index.
Doing 'db.runCommand' doesn't work either:
> db.runCommand({dropIndexes:'fs_files',index:{_id:1}})
{ "nIndexesWas" : 2, "errmsg" : "may not delete _id index", "ok" : 0 }
not ok.
Can i use a field including _id in a composite index?
I couldn't find anything online, the ensureindex command can't do it.
db.fs_files.ensureIndex({'_id':1, 'created':1});
the above command just created a new composite index. i haven't found some similar 'create Index' command.
the default _id index is a unique index?
the getIndexes returns it's not a unique index.
{
"v" : 1,
"key" : {
"_id" : 1
},
"ns" : "gridfs.fs_files",
"name" : "_id_"
},
{
"v" : 1,
"key" : {
"created" : 1
},
"unique" : true,
"ns" : "gridfs.fs_files",
"name" : "created_1"
}
There is a createIndex command in addition to ensureIndex also.
E.g.
db.<coll>.createIndex({foo:1})
You cannot delete the index on "_id" in mongodb.
Please see the documentation here
Related
I have a collection named App and need to query those active (active: true) apps that belong to a particular user (user_id) or are available to all users (by their _id). I use query like this
{
"active" : true,
"$or" : [
{
"user_id" : "111111111111111111111111"
},
{
"_id" : {
"$in" : [
ObjectId("222222222222222222222222"),
ObjectId("333333333333333333333333"),
ObjectId("444444444444444444444444")
]
}
}
]
}
However in db.currentOp(true) I see that this query is running very slowly: lockStats.timeLockedMicros.r is about 3000.
How can I optimize performance of this query? I already have the following indexes on App:
> db.App.getIndexes()
[
{
"v" : 1,
"key" : {
"_id" : 1
},
"name" : "_id_",
"ns" : "mydb.App"
},
{
"v" : 1,
"key" : {
"active" : 1,
"created_at" : -1
},
"name" : "active_1_created_at_-1",
"ns" : "mydb.App",
"background" : true
},
{
"v" : 1,
"key" : {
"active" : 1,
"user_id" : 1
},
"name" : "active_1_user_id_1",
"ns" : "mydb.App",
"background" : true
}
]
Two issues I see here:
1) You would not need index on the boolean field active as it would have low selectivity and not benefiting query performance.
"If overall selectivity is low, and if MongoDB must read a number of documents to return results, then some queries may perform faster without indexes." source
2) You need an index for user_id because user_id cannot use the compound index you created for active_1_user_id_1
Edit: You can always check index efficiency by doing a explain(true) and look at which indexes are used for that query.
I would try to do the following:
remove all your indexes, your active field has a low cardinality (boolean) and does not help you at all, you are not using created_at, so there is no reason for it.
add an index only on user_id key
change your strings as numbers to numbers.
I'm trying to add a username field to documents in a 'users' collection, and I'd like it to be a unique index. (So far, we've been using email addresses for login but we'd like to add a username field as well.) However, running db.users.ensureIndex({username:1},{unique:true}) fails because mongo considers all the unset usernames to be duplicates and therefore not unique. Anybody know how to get around this?
Show the current users and username if they have one:
> db.users.find({},{_id:0,display_name:1,username:1})
{ "display_name" : "james" }
{ "display_name" : "sammy", "username" : "sammy" }
{ "display_name" : "patrick" }
Attempt to make the 'username' field a unique index:
> db.users.ensureIndex({username:1},{unique:true})
{
"err" : "E11000 duplicate key error index: blend-db1.users.$username_1 dup key: { : null }",
"code" : 11000,
"n" : 0,
"connectionId" : 272,
"ok" : 1
}
It doesn't work because both james and sammy have username:null.
Let's set patrick's username to 'patrick' to eliminate the duplicate null value.
> db.users.update({display_name: 'patrick'}, { $set: {username: 'patrick'}});
> db.users.ensureIndex({username:1},{unique:true})
> db.users.getIndexes()
[
{
"v" : 1,
"key" : {
"_id" : 1
},
"ns" : "blend-db1.users",
"name" : "_id_"
},
{
"v" : 1,
"key" : {
"username" : 1
},
"unique" : true,
"ns" : "blend-db1.users",
"name" : "username_1"
}
]
Now it works!
To clarify the question, what I'd like is to be able to make username a unique index without having to worry about all the documents that have username still set to null.
Try creating a unique sparse index:
db.users.ensureIndex({username:1},{unique:true,sparse:true})
As per the docs:
You can combine the sparse index option with the unique indexes option
so that mongod will reject documents that have duplicate values for a
field, but that ignore documents that do not have the key.
Although this only works for documents which don't have the field, as opposed to documents that do have the field, but where the field has a null value.
I have a collection with the _id field as a IP with type String.
I'm using mongoose, but here's the error on the console:
$ db.servers.remove()
$ db.servers.insert({"_id":"1.2.3.4"})
$ db.servers.insert({"_id":"1.2.3.5"}) <-- Throws dup key: { : null }
Likely, it's because you have an index that requires a unique value for one of the fields as shown below:
> db.servers.remove()
> db.servers.ensureIndex({"name": 1}, { unique: 1})
> db.servers.insert({"_id": "1.2.3"})
> db.servers.insert({"_id": "1.2.4"})
E11000 duplicate key error index: test.servers.$name_1 dup key: { : null }
You can see your indexes using getIndexes() on the collection:
> db.servers.getIndexes()
[
{
"v" : 1,
"key" : {
"_id" : 1
},
"ns" : "test.servers",
"name" : "_id_"
},
{
"v" : 1,
"key" : {
"name" : 1
},
"unique" : true,
"ns" : "test.servers",
"name" : "name_1"
}
]
I was confused by exactly the same error today, and later figured it out. It was because I removed a indexed property from a mongoose schema, but did not drop that property from the mongodb index. The error message is infact that the new document has an indexed property whose value is null (not in the json).
What exactly happens when I call ensureIndex(data) when typical data looks like data:{name: "A",age:"B", job : "C"} ? Will it create a compound index over these three fields or will it create only one index applicable when anything from data is requested or something altogether different ?
You can do either :
> db.collection.ensureIndex({"data.name": 1,"data.age":1, "data.job" : 1})
> db.collection.ensureIndex({"data": 1})
This is discussed in the documentation under indexes-on-embedded-fields and indexes on sub documents
The important section of the sub document section is 'When performing equality matches on subdocuments, field order matters and the subdocuments must match exactly.'
This means that the 2 indexes are the same for simple queries .
However, as the sub-document example shows, you can get some interesting results (that you might not expect) if you just index the whole sub-document as opposed to a specific field and then do a comparison operator (like $gte) - if you index a specific sub field you get a less flexible, but potentially more useful index.
It really all depends on your use case.
Anyway, once you have created the index you can check what's created with :
> db.collection.getIndexes()
[
{
"v" : 1,
"key" : {
"_id" : 1
},
"ns" : "test.collection",
"name" : "_id_"
},
{
"v" : 1,
"key" : {
"data.name" : 1,
"data.age" : 1,
"data.job" : 1
},
"ns" : "test.collection",
"name" : "data.name_1_data.age_1_data.job_1"
}
]
As you can see from the output it created a new key called data.name_1_data.age_1_data.job_1 (the _id_ index is always created).
If you want to test your new index then you can do :
> db.collection.insert({data:{name: "A",age:"B", job : "C"}})
> db.collection.insert({data:{name: "A1",age:"B", job : "C"}})
> db.collection.find({"data.name" : "A"}).explain()
{
"cursor" : "BtreeCursor data.name_1_data.age_1_data.job_1",
.... more stuff
The main thing is that you can see that your new index was used (BtreeCursor data.name_1_data.age_1_data.job_1 in the cursor field is what indicates this is the case). If you see "cursor" : "BasicCursor", then your index was not used.
For more detailed information look here.
you can try this :
db.collection.ensureIndex({"data.name": 1,"data.age":1, "data.job" : 1})
When I call ensureIndex from the mongo shell on a collection for a compound index an _id field of type ObjectId is auto-generated in the index object.
> db.system.indexes.find();
{ "name" : "_id_", "ns" : "database.coll", "key" : { "_id" : 1 } }
{ "_id" : ObjectId("4ea78d66413e9b6a64c3e941"), "ns" : "database.coll", "key" : { "a.b" : 1, "a.c" : 1 }, "name" : "a.b_1_a.c_1" }
This makes intuitive sense as all documents in a collection need an _id field (even system.indexes, right?), but when I check the indexes generated by morphia's ensureIndex call for the same collection *there is no _id property*.
Looking at morphia's source code, it's clear that it's calling the same code that the shell uses, but for some reason (whether it's the fact that I'm creating a compound index or indexing an Embedded document or both) they produce different results. Can anyone explain this behavior to me?
Not exactly sure how you managed to get an _id field in the indexes collection but both shell and Morphia originated ensureIndex calls for compound indexes do not put an _id field in the index object :
> db.test.ensureIndex({'a.b':1, 'a.c':1})
> db.system.indexes.find({})
{ "v" : 1, "key" : { "_id" : 1 }, "ns" : "test.test", "name" : "_id_" }
{ "v" : 1, "key" : { "a.b" : 1, "a.c" : 1 }, "ns" : "test.test", "name" : "a.b_1_a.c_1" }
>
Upgrade to 2.x if you're running an older version to avoid running into now resolved issues. And judging from your output you are running 1.8 or earlier.